Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $k = \dfrac{a - 8}{-5a^2 + 10a + 240} \times \dfrac{-3a^2 + 9a + 54}{a - 6} $
Solution: First factor out any common factors. $k = \dfrac{a - 8}{-5(a^2 - 2a - 48)} \times \dfrac{-3(a^2 - 3a - 18)}{a - 6} $ Then factor the quadratic expressions. $k = \dfrac {a - 8} {-5(a - 8)(a + 6)} \times \dfrac {-3(a - 6)(a + 3)} {a - 6} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac {(a - 8) \times -3(a - 6)(a + 3) } { -5(a - 8)(a + 6) \times (a - 6)} $ $k = \dfrac {-3(a - 6)(a + 3)(a - 8)} {-5(a - 8)(a + 6)(a - 6)} $ Notice that $(a - 8)$ and $(a - 6)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac {-3(a - 6)(a + 3)\cancel{(a - 8)}} {-5\cancel{(a - 8)}(a + 6)(a - 6)} $ We are dividing by $a - 8$ , so $a - 8 \neq 0$ Therefore, $a \neq 8$ $k = \dfrac {-3\cancel{(a - 6)}(a + 3)\cancel{(a - 8)}} {-5\cancel{(a - 8)}(a + 6)\cancel{(a - 6)}} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $k = \dfrac {-3(a + 3)} {-5(a + 6)} $ $ k = \dfrac{3(a + 3)}{5(a + 6)}; a \neq 8; a \neq 6 $